To calculate the thrust generated by the engine, we can use the following formula for thrust in a rocket engine, which accounts for the mass flow rate of the exhaust gas and the velocity of the exhaust at the nozzle.
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First, we need to calculate the speed of sound (\(a\)) in the combustion chamber, using the equation: \[ a = \sqrt{\gamma \cdot R \cdot T} \] where:
- \(\gamma = 1.14\)
- \(R = 378.00 , \text{J/(kg·K)}\)
- \(T = 3685 , \text{K}\)
Let's calculate \(a\): \[ a = \sqrt{1.14 \cdot 378.00 \cdot 3685} \]
\[ a = \sqrt{(1.14 \cdot 378.00 \cdot 3685)} = \sqrt{1430969.92} \approx 1196.26 , \text{m/s} \]
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Next, we will calculate the exit velocity (\(v_e\)) at the nozzle exit, using the equation for isentropic flow: \[ v_e = a \cdot \sqrt{ \frac{2 \cdot \gamma}{\gamma - 1} \left( \left( \frac{P_0}{P_e} \right)^{\frac{\gamma - 1}{\gamma}} - 1 \right) } \] where:
- \(P_0 = 9.7 , \text{MPa} = 9700000 , \text{Pa}\)
- \(P_e = 101325 , \text{Pa}\)
We can now compute: \[ v_e = 1196.26 \cdot \sqrt{ \frac{2 \cdot 1.14}{1.14 - 1} \left( \left( \frac{9700000}{101325} \right)^{\frac{1.14 - 1}{1.14}} - 1 \right) } \]
We first compute: \[ \frac{9700000}{101325} \approx 95.9 \]
Then, we calculate: \[ \left( 95.9 \right)^{\frac{0.14}{1.14}} \approx 1.52 \]
So, substituting back, we find: \[ v_e = 1196.26 \cdot \sqrt{ \frac{2 \cdot 1.14}{0.14} \cdot (1.52 - 1) } \] \[ = 1196.26 \cdot \sqrt{ \approx 16.33 \cdot 0.52 } \] \[ = 1196.26 \cdot \sqrt{8.58} = 1196.26 \cdot 2.93 \approx 3504.67 , \text{m/s} \]
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Next, we calculate the mass flow rate (\( \dot{m} \)): \[ \dot{m} = \rho \cdot A \cdot v \] where:
- \(A = 0.05 , \text{m}^2\)
- \(P_0\) is the chamber pressure and we can find the density using the ideal gas law: \[ \rho = \frac{P_0}{R \cdot T_0} \]
\[ \rho = \frac{9700000}{378.00 \cdot 3685} \approx 7.092, \text{kg/m}^3 \]
Therefore: \[ \dot{m} = 7.092 \cdot 0.05 \cdot 3504.67 \approx 1247.30 , \text{kg/s} \]
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Finally, we can calculate the thrust (\(F\)): \[ F = \dot{m} \cdot v_e + (P_e - P_amb) \cdot A \]
Where \(P_{amb}\) (atmospheric pressure) is typically \(101325 , \text{Pa}\), and: \[ F \approx 1247.30 \cdot 3504.67 + (101325 - 101325) \cdot 0.05 \] \[ F \approx 4375603.91 , \text{N} \]
Finally, converting from N to kN: \[ F \approx 4375.60, \text{kN} \]
Therefore, the thrust generated by the engine is approximately 4375.6 kN.